JEE (Main) · 2024
JEE (Main) 2024 — 29 January, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Sets, Relations and Functions2
- Matrices and Determinants2
9 easy · 19 medium · 2 hard
Physics
30 Q
- Magnetic Effects of Current and Magnetism3
- Optics3
- Kinematics2
20 easy · 8 medium · 2 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques4
- Hydrocarbons3
- d- and f-Block Elements2
22 easy · 8 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctEasy | |
A easy single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions First find the values taken by $g$.
For $-3 \le x \le 0$: $g(x) = -x$ runs over $[0, 3]$.
For $0 < x \le 1$: $g(x) = x$ runs over $(0, 1]$, already inside $[0, 3]$.
So $g$ takes every value in $[0, 3]$, and $f(g(x))$ always uses the second piece of $f$: $f(t) = 1 - \frac{t}{3}$ for $0 \le t \le 3$.
As $t$ runs over $[0, 3]$, $1 - \frac{t}{3}$ decreases from 1 (at $t = 0$) to 0 (at $t = 3$), taking both end values.
Range of $f \circ g$ = $[0, 1]$.
Answer: (B). | ||||
| 2 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Reflexive: $(a, b)\,R\,(a, b)$ needs $ab - ba = 0$, which is divisible by 5. Yes.
Symmetric: if $5 \mid ad - bc$, then for $(c, d)\,R\,(a, b)$ we need $5 \mid cb - da = -(ad - bc)$. Yes.
Transitive: take $(1, 0)$, $(0, 0)$, $(0, 1)$.
$(1, 0)\,R\,(0, 0)$: $1\cdot 0 - 0\cdot 0 = 0$, divisible by 5.
$(0, 0)\,R\,(0, 1)$: $0\cdot 1 - 0\cdot 0 = 0$, divisible by 5.
But $(1, 0)\,R\,(0, 1)$ needs $1\cdot 1 - 0\cdot 0 = 1$ divisible by 5, which fails. Not transitive.
So R is reflexive and symmetric but not transitive.
Answer: (B). | ||||
| 3 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $z + 1 = \frac{3}{2} - 2i$, so $|z + 1| = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
Right side: $\alpha\left(\frac{1}{2} - 2i\right) + \beta + \beta i = \left(\frac{\alpha}{2} + \beta\right) + i(\beta - 2\alpha)$.
The left side is real, so the imaginary part is zero: $\beta = 2\alpha$.
Real part: $\frac{\alpha}{2} + 2\alpha = \frac{5}{2} \Rightarrow \frac{5\alpha}{2} = \frac{5}{2} \Rightarrow \alpha = 1$, $\beta = 2$.
$\alpha + \beta = 3$.
Answer: (C). | ||||
| 4 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Expand without assuming the matrices commute:
$(A + A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2$,
$(A - A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2$.
Adding: $2A^2 + 2(A^T)^2$.
So the expression is $\frac{1}{2}A\left[2A^2 + 2(A^T)^2\right] = A^3 + A(A^T)^2 = A^3 + (AA^T)A^T = A^3 + IA^T = A^3 + A^T$.
Answer: (B). | ||||
| 5 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $|A| = 1\cdot(\alpha^2 - \beta^2) = \alpha^2 - \beta^2$.
For a $3 \times 3$ matrix, $|2A| = 2^3|A| = 8|A|$, so $|2A|^3 = 2^9|A|^3 = 2^{21}$, giving $|A|^3 = 2^{12}$ and $|A| = 16$.
$\alpha^2 - \beta^2 = (\alpha - \beta)(\alpha + \beta) = 16$ with integers. The two factors have the same parity, so they are $(2, 8)$ or $(4, 4)$ (or the same with signs/order changed): $\alpha = \pm 5, \beta = \pm 3$, or $\alpha = \pm 4, \beta = 0$.
Among the options only $\alpha = 5$ occurs (check: $25 - 9 = 16$).
Answer: (B). | ||||
| 6 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series With common difference $d$: $a_1 = 2 - 5d$, $a_4 = 2 - 2d$, $a_5 = 2 - d$.
$P(d) = (2 - 5d)(2 - 2d)(2 - d)$. First $(2 - 5d)(2 - 2d) = 4 - 14d + 10d^2$, then
$P(d) = (4 - 14d + 10d^2)(2 - d) = 8 - 32d + 34d^2 - 10d^3$.
$P'(d) = -32 + 68d - 30d^2 = 0 \Rightarrow 15d^2 - 34d + 16 = 0 \Rightarrow d = \frac{34 \pm \sqrt{1156 - 960}}{30} = \frac{34 \pm 14}{30}$, so $d = \frac{8}{5}$ or $d = \frac{2}{3}$.
$P''(d) = 68 - 60d$: at $d = \frac{8}{5}$, $P'' = 68 - 96 < 0$ (local maximum); at $d = \frac{2}{3}$, $P'' = 28 > 0$ (local minimum).
So the product is greatest (locally) for $d = \frac{8}{5}$.
Answer: (B). | ||||
| 7 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the terms be $a, ar, ar^2, \ldots, ar^{63}$. The odd-placed terms are $a, ar^2, ar^4, \ldots$ (32 terms) with sum $S_o$.
Each even-placed term is $r$ times the odd-placed term before it, so the sum of even-placed terms is $rS_o$.
Total $= S_o + rS_o = (1 + r)S_o = 7S_o \Rightarrow r = 6$.
Answer: (C). | ||||
| 8 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability At $x = \frac{\pi}{2}$ both the integral and $\left(x - \frac{\pi}{2}\right)^2$ are 0, so use L'Hôpital's rule.
Numerator $N(x) = \int_{x^3}^{(\pi/2)^3}\cos(t^{1/3})\,dt$; by the Leibniz rule $N'(x) = -\cos\left((x^3)^{1/3}\right)\cdot 3x^2 = -3x^2\cos x$.
Denominator derivative: $2\left(x - \frac{\pi}{2}\right)$.
Limit $= \lim \frac{-3x^2\cos x}{2\left(x - \frac{\pi}{2}\right)}$. Here $\frac{\cos x}{x - \frac{\pi}{2}} \to -\sin\frac{\pi}{2} = -1$ (derivative of $\cos x$ at $\frac{\pi}{2}$).
So the limit is $\frac{-3\cdot\frac{\pi^2}{4}\cdot(-1)}{2} = \frac{3\pi^2}{8}$.
Answer: (D). | ||||
| 9 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctEasy | |
A easy single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Write $f(x) = \tan x\cdot g(x)$ with $g(x) = \frac{(2^x + 2^{-x})\sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}$, which is differentiable near 0.
$f'(x) = \sec^2 x\, g(x) + \tan x\, g'(x)$, and at $x = 0$: $f'(0) = 1\cdot g(0) + 0 = g(0)$.
$g(0) = \frac{(1 + 1)\sqrt{\tan^{-1}1}}{1^3} = 2\sqrt{\frac{\pi}{4}} = \sqrt{\pi}$.
Answer: (B). | ||||
| 10 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | StatementsMedium | |
A medium statements question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 12\sqrt{2}x^2 - 3\sqrt{2} = 3\sqrt{2}(4x^2 - 1) > 0$ for $x > \frac{1}{2}$, so $f$ is strictly increasing on $\left[\frac{1}{2}, 1\right]$.
$f\left(\frac{1}{2}\right) = \frac{\sqrt{2}}{2} - \frac{3\sqrt{2}}{2} - 1 = -\sqrt{2} - 1 < 0$ and $f(1) = \sqrt{2} - 1 > 0$. So there is exactly one zero: (I) is correct.
To find it, put $x = \cos\theta$, $\theta \in \left[0, \frac{\pi}{3}\right]$: $f = \sqrt{2}(4\cos^3\theta - 3\cos\theta) - 1 = \sqrt{2}\cos 3\theta - 1$.
$f = 0 \Rightarrow \cos 3\theta = \frac{1}{\sqrt{2}}$ with $3\theta \in [0, \pi]$, so $3\theta = \frac{\pi}{4}$, $\theta = \frac{\pi}{12}$. The zero is $x = \cos\frac{\pi}{12}$ (about 0.966, inside the interval): (II) is correct.
Answer: (A). | ||||
| 11 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Use $\int_{-a}^{a}h(x)\,dx = \int_0^a [h(x) + h(-x)]\,dx$.
First term: $\frac{x^2\cos x}{1 + \pi^x} + \frac{x^2\cos x}{1 + \pi^{-x}} = x^2\cos x\left(\frac{1}{1 + \pi^x} + \frac{\pi^x}{\pi^x + 1}\right) = x^2\cos x$.
Second term: $\sin x^{2023}$ is an odd function of $x$ (whether read as $\sin(x^{2023})$ or $(\sin x)^{2023}$), call it $u(x)$ with $u(-x) = -u(x)$; and $1 + \sin^2 x$ is even. So the pair gives $(1 + \sin^2 x)\left(\frac{1}{1 + e^{u}} + \frac{1}{1 + e^{-u}}\right) = 1 + \sin^2 x$.
So $I = \int_0^{\pi/2}x^2\cos x\,dx + \int_0^{\pi/2}(1 + \sin^2 x)\,dx$.
$\int_0^{\pi/2}x^2\cos x\,dx = \left[x^2\sin x + 2x\cos x - 2\sin x\right]_0^{\pi/2} = \frac{\pi^2}{4} - 2$.
$\int_0^{\pi/2}(1 + \sin^2 x)\,dx = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.
$I = \frac{\pi^2}{4} + \frac{3\pi}{4} - 2 = \frac{\pi}{4}(\pi + 3) - 2$, so $a = 3$.
Answer: (D). | ||||
| 12 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | Single correctHard | |
A hard single correct question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Multiply numerator and denominator by $\sin x\cos x$:
numerator $(\operatorname{cosec}x + \sin x)\sin x\cos x = \cos x(1 + \sin^2 x)$;
denominator $\operatorname{cosec}x\sec x\cdot\sin x\cos x + \tan x\sin^2 x\cdot\sin x\cos x = 1 + \sin^4 x$.
With $t = \sin x$, $dt = \cos x\,dx$: $y = \int\frac{1 + t^2}{1 + t^4}\,dt = \int\frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}}\,dt$.
Put $u = t - \frac{1}{t}$, $du = \left(1 + \frac{1}{t^2}\right)dt$, $t^2 + \frac{1}{t^2} = u^2 + 2$:
$y = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t - \frac{1}{t}}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin^2 x - 1}{\sqrt{2}\sin x}\right) + C$ (valid for $0 < x < \frac{\pi}{2}$, which contains both points used).
As $x \to \frac{\pi}{2}^-$, $\sin x \to 1$ and the argument $\to 0$, so $0 = 0 + C$, $C = 0$.
At $x = \frac{\pi}{4}$: $\sin x = \frac{1}{\sqrt{2}}$, $\frac{\frac{1}{2} - 1}{\sqrt{2}\cdot\frac{1}{\sqrt{2}}} = -\frac{1}{2}$.
$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(-\frac{1}{2}\right)$.
Answer: (B). | ||||
| 13 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Notice that $\frac{d}{dx}(1 + \cos^2 x) = -2\cos x\sin x = -\sin 2x$. So
$\frac{d}{dx}\left[f(x)(1 + \cos^2 x)\right] = f'(x)(1 + \cos^2 x) - f(x)\sin 2x$.
The equation says exactly that this equals $\sin x$:
$\frac{d}{dx}\left[f(x)(1 + \cos^2 x)\right] = \sin x \Rightarrow f(x)(1 + \cos^2 x) = -\cos x + C$.
$f(0) = 0$: $0 = -1 + C$, so $C = 1$ and $f(x) = \frac{1 - \cos x}{1 + \cos^2 x}$.
$f\left(\frac{\pi}{2}\right) = \frac{1 - 0}{1 + 0} = 1$.
(Equivalently: the linear form $f' - \frac{\sin 2x}{1 + \cos^2 x}f = \frac{\sin x}{1 + \cos^2 x}$ has integrating factor $1 + \cos^2 x$.)
Answer: (B). | ||||
| 14 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines AB is vertical (same x) and AC is horizontal (same y), so the angle at A is a right angle. The circumcentre of a right triangle is the midpoint of the hypotenuse BC:
$\left(\frac{a + \frac{a}{4}}{2}, \frac{6 - 2}{2}\right) = \left(\frac{5a}{8}, 2\right)$.
Equate with $\left(5, \frac{a}{4}\right)$: $\frac{a}{4} = 2 \Rightarrow a = 8$, and $\frac{5a}{8} = 5$ agrees.
Vertices $A(8, -2)$, $B(8, 6)$, $C(2, -2)$: $AB = 8$, $AC = 6$, $BC = 10$.
$\alpha = \frac{BC}{2} = 5$, $\beta = \frac{1}{2}\cdot 8\cdot 6 = 24$, $\gamma = 8 + 6 + 10 = 24$.
$\alpha + \beta + \gamma = 53$.
Answer: (D). | ||||
| 15 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | Single correctHard | |
A hard single correct question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines The bisector of angle B meets AC at D: $y = x$ and $2x - y = 2$ give $D(2, 2)$.
Angle-bisector theorem: $\frac{AD}{DC} = \frac{AB}{BC} = \frac{1}{2}$, so $DC = 2AD$ and C lies on line AC beyond D: $C = D + 2(D - A) = (2, 2) + 2(-2, -4) = (-2, -6)$. (Check: $2(-2) - (-6) = 2$, on AC.)
The image of A in the bisector $y = x$ is $A'(6, 4)$; it lies on line BC.
Line through $C(-2, -6)$ and $A'(6, 4)$: slope $\frac{10}{8} = \frac{5}{4}$, so $y + 6 = \frac{5}{4}(x + 2)$.
B is where this meets $y = x$: $x + 6 = \frac{5}{4}x + \frac{5}{2} \Rightarrow \frac{x}{4} = \frac{7}{2} \Rightarrow x = 14$. So $B(14, 14)$.
Check: $AB^2 = 100 + 64 = 164$, $BC^2 = 256 + 400 = 656 = 4\cdot 164$, so $BC = 2AB$.
$\alpha + 2\beta = 14 + 28 = 42$.
Answer: (B). | ||||
| 16 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Centroid $G = \frac{P + Q + R}{3} = \frac{2M + R}{3} = \frac{(4, 2, 4) + (-1, 4, 2)}{3} = (1, 2, 2)$.
First line: $(2, 2s, -3 - s)$. Second line: $(1 + t, -3 - 3t, -1 + t)$.
x: $2 = 1 + t \Rightarrow t = 1$, giving the point $(2, -6, 0)$ on the second line.
On the first line, $2s = -6 \Rightarrow s = -3$, $z = -3 + 3 = 0$. Same point, so the lines meet at $(2, -6, 0)$.
Distance from $G$: $\sqrt{1^2 + 8^2 + 2^2} = \sqrt{69}$.
Answer: (B). | ||||
| 17 | Vector AlgebraAddition of vectors | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on addition of vectors, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a} + 5\vec{b} = \lambda\vec{c}$ and $\vec{b} + 6\vec{c} = \mu\vec{a}$.
From the first, $\vec{a} = \lambda\vec{c} - 5\vec{b}$. Put into the second: $\vec{b} + 6\vec{c} = \mu\lambda\vec{c} - 5\mu\vec{b}$, i.e. $(1 + 5\mu)\vec{b} + (6 - \lambda\mu)\vec{c} = \vec{0}$.
Since $\vec{b}, \vec{c}$ are non-collinear, both coefficients vanish: $\mu = -\frac{1}{5}$, $\lambda\mu = 6 \Rightarrow \lambda = -30$.
So $\vec{a} + 5\vec{b} + 30\vec{c} = \vec{0}$, and (because $\vec{b}, \vec{c}$ are independent the representation is unique) $\alpha = 5$, $\beta = 30$.
$\alpha + \beta = 35$.
Answer: (D). | ||||
| 18 | Vector AlgebraAddition of vectors | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on addition of vectors, from Vector Algebra. NCERT: Class 12: Vector Algebra $|OA| = \sqrt{4 + 4 + 1} = 3$, $|OB| = \sqrt{4 + 16 + 16} = 6$.
The internal bisector from O divides AB in the ratio $AC : CB = OA : OB = 1 : 2$.
$\vec{OC} = \frac{2\vec{OA} + 1\cdot\vec{OB}}{3} = \frac{(4, 4, 2) + (2, 4, 4)}{3} = \left(2, \frac{8}{3}, 2\right)$.
$|OC| = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{\frac{136}{9}} = \frac{2\sqrt{34}}{3}$.
Answer: (A). | ||||
| 19 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability P(first 2 on throw $k$) $= \left(\frac{5}{6}\right)^{k-1}\frac{1}{6}$ (independent throws).
P(even $k$) $= \frac{1}{6}\left[\frac{5}{6} + \left(\frac{5}{6}\right)^3 + \left(\frac{5}{6}\right)^5 + \cdots\right] = \frac{1}{6}\cdot\frac{\frac{5}{6}}{1 - \frac{25}{36}} = \frac{1}{6}\cdot\frac{5}{6}\cdot\frac{36}{11} = \frac{5}{11}$.
Check: P(odd) $= \frac{1}{6}\cdot\frac{1}{1 - \frac{25}{36}} = \frac{6}{11}$, and $\frac{5}{11} + \frac{6}{11} = 1$.
Answer: (D). | ||||
| 20 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions In $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\cos\alpha > 0$. Divide by $\cos\alpha$ and write $T = \tan\alpha$: $4 + 5T = \sec\alpha = \sqrt{1 + T^2}$ (so we need $4 + 5T > 0$).
Square: $16 + 40T + 25T^2 = 1 + T^2 \Rightarrow 24T^2 + 40T + 15 = 0$.
$T = \frac{-40 \pm \sqrt{1600 - 1440}}{48} = \frac{-40 \pm 4\sqrt{10}}{48} = \frac{-10 \pm \sqrt{10}}{12}$.
$T = \frac{-10 - \sqrt{10}}{12} \approx -1.10$ gives $4 + 5T < 0$: extraneous. $T = \frac{\sqrt{10} - 10}{12} \approx -0.57$ gives $4 + 5T \approx 1.15 > 0$: valid.
$\tan\alpha = \frac{\sqrt{10} - 10}{12}$.
Answer: (C). | ||||
| 21 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Each root satisfies $x^2 = x - 2$. Reduce powers:
$x^4 = (x - 2)^2 = x^2 - 4x + 4 = (x - 2) - 4x + 4 = -3x + 2$.
$x^6 = x^2\cdot x^4 = (x - 2)(-3x + 2) = -3x^2 + 8x - 4 = -3(x - 2) + 8x - 4 = 5x + 2$.
So $\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2 = (5\alpha + 2) + (-3\alpha + 2) + (-3\beta + 2) - 5(\alpha - 2)$
$= -3\alpha - 3\beta + 16 = -3(\alpha + \beta) + 16 = -3(1) + 16 = 13$.
(The condition on imaginary parts does not change the value.)
Answer: 13. | ||||
| 22 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | NumericalMedium | |
A medium numerical question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations Letters: E, G, N, T, T, W, Y (T twice), in alphabetical order E < G < N < T < W < Y.
Words before GTWENTY:
1st letter E: remaining G, N, T, T, W, Y: $\frac{6!}{2!} = 360$.
Starting with G, 2nd letter E or N (before T): each leaves 5 letters with T twice: $2\times\frac{5!}{2!} = 120$.
Starting GT, remaining E, N, T, W, Y; 3rd letter E, N or T (before W): $3\times 4! = 72$.
Starting GTW, remaining E, N, T, Y; 4th letter E is the smallest: 0.
GTWE, remaining N, T, Y; N is smallest: 0. GTWEN, remaining T, Y; T is smallest: 0. Then GTWENTY itself.
Rank $= 360 + 120 + 72 + 1 = 553$.
Answer: 553. | ||||
| 23 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Use $\frac{{}^{11}C_r}{r + 1} = \frac{11!}{(r + 1)!\,(11 - r)!} = \frac{1}{12}\cdot{}^{12}C_{r+1}$.
Sum $= \frac{1}{12}\left({}^{12}C_2 + {}^{12}C_3 + \cdots + {}^{12}C_{10}\right)$.
${}^{12}C_2 + \cdots + {}^{12}C_{10} = 2^{12} - ({}^{12}C_0 + {}^{12}C_1 + {}^{12}C_{11} + {}^{12}C_{12}) = 4096 - (1 + 12 + 12 + 1) = 4070$.
Sum $= \frac{4070}{12} = \frac{2035}{6}$; $2035 = 5\cdot 11\cdot 37$ shares no factor with 6.
$n + m = 2035 + 6 = 2041$.
Answer: 2041. | ||||
| 24 | Applications of DerivativesIncreasing and decreasing functions | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on increasing and decreasing functions, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Zeros of $f$: $f(2) = 4 - 4 = 0$, $f(4) = 16 - 16 = 0$, and $f(-1) = \frac{1}{2} - 1 < 0$, $f(0) = 1 > 0$, so there is a zero in $(-1, 0)$. That is at least 3.
$f''(x) = 2^x(\ln 2)^2 - 2$ is strictly increasing, so it has exactly one zero. Then $f'$ has at most 2 zeros and $f$ at most 3 zeros (Rolle's theorem each time).
So $m = 3$. By Rolle between consecutive zeros of $f$, $f'$ has at least 2 zeros, and at most 2: $n = 2$.
(Directly: $f'(x) = 2^x\ln 2 - 2x$; $f'(0) = \ln 2 > 0$, $f'(2) = 4\ln 2 - 4 < 0$, $f'(5) = 32\ln 2 - 10 > 0$: two sign changes.)
$m + n = 5$.
Answer: 5. | ||||
| 25 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals The line is $y = 5x - 13$. Meeting the circle: $x^2 + (5x - 13)^2 = 169 \Rightarrow 26x^2 - 130x = 0$, so $x = 0$ or $x = 5$: points $P(0, -13)$ and $Q(5, 12)$.
The origin gives $0 < 5\cdot 0 - 13$? No, so the centre lies above the line; the part below the line is the minor segment cut off by chord PQ.
Central angle $\theta$ between OP and OQ: $\cos\theta = \frac{0\cdot 5 + (-13)(12)}{169} = -\frac{12}{13}$, $\sin\theta = \frac{5}{13}$.
$\theta = \pi - \cos^{-1}\frac{12}{13} = \pi - \left(\frac{\pi}{2} - \sin^{-1}\frac{12}{13}\right) = \frac{\pi}{2} + \sin^{-1}\frac{12}{13}$.
Segment area $= \frac{r^2}{2}(\theta - \sin\theta) = \frac{169}{2}\left(\frac{\pi}{2} + \sin^{-1}\frac{12}{13}\right) - \frac{169}{2}\cdot\frac{5}{13} = \frac{169\pi}{4} - \frac{65}{2} + \frac{169}{2}\sin^{-1}\frac{12}{13}$.
Matching: $\frac{\alpha}{\beta} = \frac{169}{2}$ (and $\frac{\pi\alpha}{2\beta} = \frac{169\pi}{4}$ agrees), so $\alpha = 169$, $\beta = 2$.
$\alpha + \beta = 171$.
Answer: 171. | ||||
| 26 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalEasy | |
A easy numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Separate: $\frac{dy}{1 + y^2} = -\frac{1 + \ln x}{x}\,dx$.
Integrate: $\tan^{-1}y = -\left(\ln x + \frac{(\ln x)^2}{2}\right) + C$.
At $(1, 1)$: $\frac{\pi}{4} = 0 + C$.
At $x = e$: $\tan^{-1}y = \frac{\pi}{4} - \left(1 + \frac{1}{2}\right) = \frac{\pi}{4} - \frac{3}{2}$.
$y(e) = \tan\left(\frac{\pi}{4} - \frac{3}{2}\right) = \frac{1 - \tan\frac{3}{2}}{1 + \tan\frac{3}{2}}$.
So $\alpha = 1$, $\beta = 1$ and $\alpha + 2\beta = 3$.
Answer: 3. | ||||
| 27 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Centre = meet of the diameters: $2x - 3y = 5$, $3x - 4y = 7$. Multiply to $6x - 9y = 15$ and $6x - 8y = 14$; subtract: $y = -1$, then $x = 1$. Centre $O(1, -1)$.
The line through the two points: slope $\frac{3 - (-4)}{-\frac{1}{7} + \frac{22}{7}} = \frac{7}{3}$; $y + 4 = \frac{7}{3}\left(x + \frac{22}{7}\right) \Rightarrow 7x - 3y + 10 = 0$.
It meets the circle at one point, so it is a tangent and P is the foot of the perpendicular from the centre.
$7(1) - 3(-1) + 10 = 20$, and $7^2 + 3^2 = 58$.
$P = (1, -1) - \frac{20}{58}(7, -3) = \left(1 - \frac{70}{29}, -1 + \frac{30}{29}\right) = \left(-\frac{41}{29}, \frac{1}{29}\right)$.
Check: $7\left(-\frac{41}{29}\right) - \frac{3}{29} + 10 = \frac{-287 - 3 + 290}{29} = 0$.
$17\beta - \alpha = \frac{17}{29} + \frac{41}{29} = \frac{58}{29} = 2$.
Answer: 2. | ||||
| 28 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections On $y^2 = 3x^2$ the circle gives $x^2 + 3x^2 = 4b$, so $x^2 = b$ and $y^2 = 3b$.
These points are on the ellipse: $\frac{b}{16} + \frac{3b}{b^2} = 1 \Rightarrow \frac{b}{16} + \frac{3}{b} = 1 \Rightarrow b^2 - 16b + 48 = 0 \Rightarrow b = 4$ or $b = 12$.
$b = 4$ makes the "ellipse" $\frac{x^2}{16} + \frac{y^2}{16} = 1$, the same circle $x^2 + y^2 = 16$: not distinct. So $b = 12$.
Then $x^2 = 12$, $y^2 = 36$: the points $(\pm 2\sqrt{3}, \pm 6)$, a rectangle $4\sqrt{3}$ by 12, area $48\sqrt{3}$.
$3\sqrt{3}\times 48\sqrt{3} = 432$.
Answer: 432. | ||||
| 29 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry First line: $P = (t, t - 2, t)$. Second line ($x + 2 = 2y = 2z = 2s$): $Q = (2s - 2, s, s)$.
$\vec{PQ} = (2s - 2 - t,\ s - t + 2,\ s - t)$ must be proportional to $(2, 1, 2)$.
x and z components equal (both have ratio 2): $2s - 2 - t = s - t \Rightarrow s = 2$, so $Q = (2, 2, 2)$.
Then $\vec{PQ} = (2 - t, 4 - t, 2 - t)$ and $2 - t = 2(4 - t) \Rightarrow t = 6$, $P = (6, 4, 6)$; $\vec{PQ} = (-4, -2, -4)$, parallel to $(2, 1, 2)$.
From $A(1, 2, 12)$: $\vec{QA} = (-1, 0, 10)$, $|\vec{QA}|^2 = 101$.
Projection on the unit direction $\frac{(2, 1, 2)}{3}$: $\frac{-2 + 0 + 20}{3} = 6$.
$l^2 = 101 - 36 = 65$.
Check by cross product: $(-1, 0, 10)\times(2, 1, 2) = (0\cdot 2 - 10\cdot 1,\ 10\cdot 2 - (-1)\cdot 2,\ -1\cdot 1 - 0) = (-10, 22, -1)$, $|\cdot|^2 = 585$, divided by $|(2,1,2)|^2 = 9$: 65.
Answer: 65. | ||||
| 30 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalEasy | |
A easy numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Ten values. Mean 56: sum = 560. The known eight add to $65 + 68 + 58 + 44 + 48 + 45 + 60 + 60 = 448$, so $\alpha + \beta = 112$.
Variance $= \frac{\sum x^2}{10} - 56^2 = 66.2 \Rightarrow \sum x^2 = 10(66.2 + 3136) = 32022$.
Known squares: $4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + 3600 = 25678$.
$\alpha^2 + \beta^2 = 32022 - 25678 = 6344$.
Check: $2\alpha\beta = 112^2 - 6344 = 6200$, so $\alpha, \beta$ are roots of $t^2 - 112t + 3100 = 0$: $t = 62, 50$, real and with $\alpha = 62 > \beta = 50$.
Answer: 6344. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement For a quotient the fractional errors add: $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
$\frac{5}{200} = 2.5\%$ and $\frac{0.2}{20} = 1\%$.
Percentage error in R $= 2.5\% + 1\% = 3.5\%$.
Answer: (C). | ||||
| 32 | KinematicsMotion in a straight line | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion in a straight line, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane From rest, $S = \frac{1}{2}at^2$.
$S_1 = \frac{1}{2}a(p - 1)^2$, $S_2 = \frac{1}{2}ap^2$.
Time $t$ for $S_1 + S_2$: $\frac{1}{2}at^2 = \frac{1}{2}a\left[(p - 1)^2 + p^2\right] \Rightarrow t^2 = 2p^2 - 2p + 1$.
$t = \sqrt{2p^2 - 2p + 1}$ s.
Answer: (A). | ||||
| 33 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion Centripetal force $F = \frac{mv^2}{r}$. Same $m$ and same $F$ means $v^2 \propto r$, so $v \propto \sqrt{r}$.
$\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Answer: (C) $\sqrt{3} : 2$. | ||||
| 34 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion On a horizontal surface the normal force is $mg$, so kinetic friction is $f = \mu mg = 0.4\times 100\times 10 = 400$ N (taking $g = 10$ m s$^{-2}$).
Work done against friction $= f\times d = 400\times 10 = 4000$ J.
Answer: (A). | ||||
| 35 | Work, Energy and PowerConservation of energy | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on conservation of energy, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power For a conservative force, $F_x = -\frac{\partial U}{\partial x} = -4x$.
At $P(1, 2, 3)$: $F_x = -4$ N, magnitude 4 N.
Answer: (B). | ||||
| 36 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At height $h$: $g_h = g\frac{R^2}{(R + h)^2}$. At depth $h$: $g_d = g\left(1 - \frac{h}{R}\right)$.
Equal: $\frac{R^2}{(R + h)^2} = \frac{R - h}{R} \Rightarrow R^3 = (R - h)(R + h)^2$.
With $x = \frac{h}{R}$: $1 = (1 - x)(1 + x)^2 = 1 + x - x^2 - x^3 \Rightarrow x(1 - x - x^2) = 0$.
$x \neq 0$, so $x^2 + x - 1 = 0 \Rightarrow x = \frac{\sqrt{5} - 1}{2}$.
$h = \frac{\sqrt{5}R - R}{2}$.
Answer: (C). | ||||
| 37 | Properties of Bulk MatterSurface tension | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | StatementsEasy | |
A easy statements question on surface tension, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Capillary rise $h = \frac{2T\cos\theta}{r\rho g}$.
The surface tension of water falls as temperature rises (much faster than the small fall in density), so $h$ is smaller in hot water.
Statement I is true; Statement II, which says the rise is smaller in cold water, is false.
Answer: (C). | ||||
| 38 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctMedium | |
A medium single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Read off the figure: A at (3 m$^3$, 8000 dyne/cm$^2$), B at (7 m$^3$, 4000 dyne/cm$^2$), C at (3 m$^3$, 4000 dyne/cm$^2$).
Convert pressure: 1 dyne/cm$^2$ = $\frac{10^{-5}\text{ N}}{10^{-4}\text{ m}^2}$ = 0.1 Pa, so 8000 dyne/cm$^2$ = 800 Pa and 4000 dyne/cm$^2$ = 400 Pa.
A → B (expansion): work = area of the trapezium under the line $= \frac{800 + 400}{2}\times(7 - 3) = 2400$ J.
B → C (isobaric compression at 400 Pa): work $= 400\times(3 - 7) = -1600$ J.
Total work by the gas $= 2400 - 1600 = 800$ J, the area of triangle ABC ($\frac{1}{2}\times 4\times 400$).
800 J is not among the options. (Even keeping the pressure in dyne/cm$^2$ the total is 8000, which is not an option either.)
No option is correct. | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryIdeal gas equation | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on ideal gas equation, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $PV = nRT$ with the same V and T, so $P \propto n$.
$n_{H_2} = \frac{1}{2}$ mol, $n_{O_2} = \frac{1}{32}$ mol.
$\frac{P_A}{P_B} = \frac{1/2}{1/32} = 16$.
Answer: (B). | ||||
| 40 | ElectrostaticsGauss's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctEasy | |
A easy single correct question on gauss's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance $5Q$ is at distance $3a < 4a$ from the centre: inside the sphere.
$-2Q$ is at distance $5a > 4a$: outside, so it contributes no net flux.
Gauss's law: $\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{5Q}{\varepsilon_0}$.
Answer: (C). | ||||
| 41 | Current ElectricityDrift velocity and mobility | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on drift velocity and mobility, from Current Electricity. NCERT: Class 12: Current Electricity $q = \int_0^{20}I\,dt = \int_0^{20}(20 + 3t)\,dt = \left[20t + \frac{3t^2}{2}\right]_0^{20} = 400 + 600 = 1000$ C.
Answer: (A). | ||||
| 42 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The shunt is in parallel with the coil: $I_gG = (I - I_g)S$.
$S = \frac{I_gG}{I - I_g} = \frac{3\times 10^{-3}\times 10}{8 - 0.003} = \frac{0.03}{7.997} \approx 3.75\times 10^{-3}\,\Omega$.
Answer: (C). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Energy is conserved between the capacitor and the inductor: $\frac{1}{2}CV^2 = \frac{1}{2}LI_{\max}^2$.
$I_{\max} = V\sqrt{\frac{C}{L}} = 12\sqrt{\frac{100\times 10^{-6}}{6.4\times 10^{-3}}} = 12\sqrt{\frac{1}{64}} = \frac{12}{8} = 1.5$ A.
Answer: (C). | ||||
| 44 | Electromagnetic WavesDisplacement current | Class 12: Electromagnetic Waves | Matching listEasy | |
A easy matching list question on displacement current, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves A: the line integral of B equals $\mu_0$ times conduction plus displacement current: Ampere–Maxwell law (IV).
B: the line integral of E (the induced emf) equals the rate of change of magnetic flux: Faraday's law (III). (The usual form carries a minus sign, Lenz's law; the paper prints it without.)
C: the flux of E through a closed surface is $\frac{Q}{\varepsilon_0}$: Gauss's law for electricity (I).
D: the flux of B through a closed surface is zero: Gauss's law for magnetism (II).
A→IV, B→III, C→I, D→II.
Answer: (B). | ||||
| 45 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Lens maker's formula: $\frac{1}{f} = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
In air: $\frac{1}{20} = 0.5\,K$, so $K = \frac{1}{10}$ cm$^{-1}$.
In the liquid: $\frac{1}{f'} = \left(\frac{1.5}{1.6} - 1\right)\frac{1}{10} = -\frac{0.1}{1.6}\cdot\frac{1}{10} = -\frac{1}{160}$.
$f' = -160$ cm (the lens becomes diverging).
Answer: (A). | ||||
| 46 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctMedium | |
A medium single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter Same wavelength means the same momentum $p = \frac{h}{\lambda}$.
Electron (non-relativistic): $K_e = \frac{p^2}{2m} = \frac{pv}{2}$. Photon: $E = pc$.
$\frac{K_e}{E} = \frac{v}{2c} = \frac{0.25}{2} = \frac{1}{8}$.
Answer: (A). | ||||
| 47 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Single correctMedium | |
A medium single correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Add the two steps: the neutron and the tritium cancel, leaving ${}_3Li^6 + {}_1H^2 \rightarrow 2\,{}_2He^4$.
Mass defect: $6.01690 + 2.01471 - 2\times 4.00388 = 8.03161 - 8.00776 = 0.02385$ amu.
Energy $= 0.02385\times 931.5 \approx 22.22$ MeV.
Answer: (C). | ||||
| 48 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctEasy | |
A easy single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits With 10 V supplied, the Zener (reverse-biased) breaks down and holds 3 V across A–B, which is also the voltage across the 2 kΩ load (C–D).
Current through the 1 kΩ series resistor: $\frac{10 - 3}{1000} = 7$ mA.
Load current: $\frac{3}{2000} = 1.5$ mA.
$I_z = 7 - 1.5 = 5.5$ mA.
Answer: (D). | ||||
| 49 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctEasy | |
A easy single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Convex mirror: $f = +\frac{R}{2} = +15$ cm (sign convention: distances measured from the pole, positive behind the mirror).
A convex mirror forms a virtual, erect image: $m = -\frac{v}{u} = +\frac{1}{2} \Rightarrow v = -\frac{u}{2}$.
Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow -\frac{2}{u} + \frac{1}{u} = \frac{1}{15} \Rightarrow -\frac{1}{u} = \frac{1}{15} \Rightarrow u = -15$ cm.
Check: $v = 7.5$ cm, $\frac{1}{7.5} - \frac{1}{15} = \frac{1}{15}$.
Answer: (C) $-15$ cm. | ||||
| 50 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Deflection is proportional to the current through the coil. With the same total current I, the coil current falls to $I_g = \frac{5}{25}I = \frac{I}{5}$.
The rest, $\frac{4I}{5}$, goes through the shunt. Same voltage across both: $I_gG = (I - I_g)S \Rightarrow \frac{I}{5}G = \frac{4I}{5}\times 24$.
$G = 96\,\Omega$.
Answer: (B). | ||||
| 51 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalMedium | |
A medium numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane To land on step 5 the ball must just clear the outer edge of step 4, which is 0.4 m out and 0.4 m down from the starting point.
Time to fall 0.4 m: $t = \sqrt{\frac{2\times 0.4}{10}} = \sqrt{0.08}$ s.
Horizontal: $u t = 0.4 \Rightarrow u = \frac{0.4}{\sqrt{0.08}} = \sqrt{\frac{0.16}{0.08}} = \sqrt{2}$ m/s.
(Any slower and it lands on step 4 or earlier; the parabola $y = \frac{5x^2}{u^2}$ with $u^2 = 2$ gives $y = 2.5x^2$, which stays above every earlier step corner $y = x$ for $x < 0.4$.)
$x = 2$.
Answer: 2. | ||||
| 52 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | NumericalEasy | |
A easy numerical question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Rolling without slipping: $a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$. For a solid cylinder $\frac{I}{mR^2} = \frac{1}{2}$.
$a = \frac{2}{3}g\sin 60^\circ = \frac{2}{3}\times 10\times\frac{\sqrt{3}}{2} = \frac{10}{\sqrt{3}}$ m/s$^2$.
$x = 10$.
Answer: 10. | ||||
| 53 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Bernoulli: the pressure difference between the lower and upper surfaces is $\Delta P = \frac{1}{2}\rho(v_1^2 - v_2^2)$.
$\Delta P = \frac{1}{2}\times 1.2\times(4900 - 4225) = 0.6\times 675 = 405$ Pa.
Lift $= \Delta P\times A = 405\times 2 = 810$ N.
Answer: 810. | ||||
| 54 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Total energy $E = \frac{1}{2}kA^2$; kinetic energy at displacement y is $K = \frac{1}{2}k(A^2 - y^2)$.
With $y = \frac{A}{3}$: $K = \frac{1}{2}kA^2\left(1 - \frac{1}{9}\right) = \frac{8}{9}E$.
$\frac{E}{K} = \frac{9}{8}$, so $x = 9$.
Answer: 9. | ||||
| 55 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalHard | |
A hard numerical question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance The sheet's field is $E = \frac{\sigma}{2\epsilon_0}$, uniform. The electron is attracted towards the sheet with constant acceleration $a = \frac{eE}{m} = \frac{e\sigma}{2m\epsilon_0}$.
Measure displacement towards the sheet. If the initial velocity has component $u_\perp$ towards the sheet ($-1 \le u_\perp \le 1$ m/s), reaching it exactly at $t = 1$ s needs $1 = u_\perp(1) + \frac{1}{2}a(1)^2$, so $a = 2(1 - u_\perp)$.
$a$ (and so $\sigma$) is largest when the electron starts moving directly away from the sheet, $u_\perp = -1$: $a = 4$ m/s$^2$.
$\sigma = \frac{2m\epsilon_0 a}{e} = 8\,\frac{m\epsilon_0}{e}$.
$\alpha = 8$.
Answer: 8. | ||||
| 56 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity Each side is $4\,\Omega$. Battery across side AB of square ABCD: AB ($4\,\Omega$) is in parallel with A–D–C–B ($12\,\Omega$): $\frac{4\times 12}{16} = 3\,\Omega$.
In steady state the capacitor carries no current. Total current $I = \frac{9}{3 + 1} = \frac{9}{4}$ A; terminal voltage $V_{AB} = \frac{9}{4}\times 3 = \frac{27}{4}$ V.
Take $V_B = 0$, $V_A = \frac{27}{4}$. Current along A–D–C–B: $\frac{27/4}{12} = \frac{9}{16}$ A, a drop of $\frac{9}{4}$ V per side: $V_D = \frac{18}{4} = 4.5$ V, $V_C = \frac{9}{4}$ V.
Diagonal AC: $\frac{27}{4} - \frac{9}{4} = 4.5$ V; diagonal BD: $4.5 - 0 = 4.5$ V. Either diagonal gives 4.5 V.
$U = \frac{1}{2}CV^2 = \frac{1}{2}\times 4\times 4.5^2 = 40.5\,\mu$J $= \frac{81}{2}\,\mu$J.
$x = 81$.
Answer: 81. | ||||
| 57 | Magnetic Effects of Current and MagnetismMagnetic dipole | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on magnetic dipole, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter On the axis of a short magnetic dipole, the magnetic scalar potential is $V = \frac{\mu_0}{4\pi}\frac{M}{r^2}$.
$M = \frac{Vr^2}{\mu_0/4\pi} = \frac{1.5\times 10^{-5}\times(0.2)^2}{10^{-7}} = \frac{1.5\times 10^{-5}\times 0.04}{10^{-7}} = 6$ A m$^2$.
Answer: 6. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalMedium | |
A medium numerical question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current A loop in the vertical east–west plane has its normal along north–south. A horizontal field towards the north-east makes $45^\circ$ with that normal.
Initial flux $\Phi = BA\cos 45^\circ = 0.20\times(0.1)^2\times\frac{1}{\sqrt{2}} = \frac{2\times 10^{-3}}{\sqrt{2}} = \sqrt{2}\times 10^{-3}$ Wb.
It falls to zero uniformly in 1 s: $|\varepsilon| = \frac{\sqrt{2}\times 10^{-3}}{1} = \sqrt{2}\times 10^{-3}$ V.
$x = 2$.
Answer: 2. | ||||
| 59 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalHard | |
A hard numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics From the figure: the source and P lie on the line through $S_2$, each at distance D from the slit plane on opposite sides, and $S_1$ is a distance d above $S_2$.
Path via $S_2$: $D + D = 2D$. Path via $S_1$: $2\sqrt{D^2 + d^2} \approx 2D + \frac{d^2}{D}$.
Path difference $\Delta = 2\left(\sqrt{D^2 + d^2} - D\right) \approx \frac{d^2}{D}$.
Dark fringe, least d: $\Delta = \frac{\lambda}{2} \Rightarrow d^2 = \frac{\lambda D}{2} = \frac{400\times 10^{-9}\times 0.2}{2} = 4\times 10^{-8}$ m$^2$.
$d = 2\times 10^{-4}$ m $= 0.2$ mm.
Answer: 0.2. | ||||
| 60 | Atoms and NucleiHydrogen spectrum | Class 12: Atoms; Class 12: Nuclei | NumericalMedium | |
A medium numerical question on hydrogen spectrum, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Photon energy: $E = 13.6\left(1 - \frac{1}{4}\right) = 10.2$ eV $= 10.2\times 1.6\times 10^{-19}$ J.
Photon momentum $p = \frac{E}{c} = \frac{10.2\times 1.6\times 10^{-19}}{3\times 10^8} = 5.44\times 10^{-27}$ kg m/s.
Momentum conservation: the atom recoils with the same momentum, $v = \frac{p}{m} = \frac{5.44\times 10^{-27}}{1.6\times 10^{-27}} = 3.4$ m/s.
$\frac{x}{5} = 3.4 \Rightarrow x = 17$.
Answer: 17. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Structure of AtomQuantum numbers | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on quantum numbers, from Structure of Atom. NCERT: Class 11: Structure of Atom Rb: [Kr] 5s$^1$. The valence electron is in 5s: $n = 5$, $l = 0$, and then $m_l$ can only be 0; $s = +\frac{1}{2}$ (or $-\frac{1}{2}$).
(B) and (C) have $l = 1$ (a 5p electron); (D) has $m_l = 1$ with $l = 0$, which is impossible.
Answer: (A) $5, 0, 0, +\frac{1}{2}$. | ||||
| 62 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics At constant T and P: $\Delta G < 0$ for a spontaneous process, $\Delta G > 0$ for a non-spontaneous one, and $\Delta G = 0$ for a reversible process (equilibrium).
So (A), (C) and (D) are correct statements; (B), "$\Delta G$ is positive for a spontaneous reaction", is wrong.
Answer: (B). | ||||
| 63 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions Cl goes from 0 to +1 in $\mathrm{ClO^-}$ (loses 1 e$^-$) and from 0 to $-1$ in $\mathrm{Cl^-}$ (gains 1 e$^-$), so $\mathrm{ClO^-}$ and $\mathrm{Cl^-}$ form in a 1 : 1 ratio.
$\mathrm{Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O}$.
Check: Cl 2 = 2; O 2 = 1 + 1; H 2 = 2; charge $-2 = -1 - 1$.
$a, b, c, d = 1, 2, 1, 1$.
Answer: (C). | ||||
| 64 | Classification of Elements and PeriodicityIonization enthalpy | Class 11: Classification of Elements and Periodicity in Properties | Assertion and reasonEasy | |
A easy assertion and reason question on ionization enthalpy, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Across a period electrons enter the same shell, which shields poorly; the nuclear charge rises by one each step and outweighs the small extra shielding. R is true.
So the effective nuclear charge increases, the atom gets smaller and the first ionisation enthalpy generally increases across a period (with small dips at groups 13 and 16). A, which says it decreases, is false.
Answer: (D). | ||||
| 65 | Isolation of MetalsOres and concentration | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on ores and concentration, from Isolation of Metals. NCERT: No matching NCERT chapter Cryolite is $\mathrm{Na_3AlF_6}$: correct.
Carnallite is $\mathrm{KCl\cdot MgCl_2\cdot 6H_2O}$: correct.
Fluoroapatite is $\mathrm{3Ca_3(PO_4)_2\cdot CaF_2}$: correct.
Fluorspar (fluorite) is $\mathrm{CaF_2}$, not $\mathrm{BF_3}$ ($\mathrm{BF_3}$ is a gas, not a mineral). This pair is incorrect.
Answer: (C). | ||||
| 66 | p-Block ElementsGroup 14 elements | No matching NCERT chapter | StatementsMedium | |
A medium statements question on group 14 elements, from p-Block Elements. NCERT: No matching NCERT chapter Statement I: electronegativity falls from C (2.5) to Si (1.8), but from Si to Pb it does not keep falling: Si 1.8, Ge 1.8, Sn 1.8, Pb 1.9 (Pauling values) — roughly constant, Pb even slightly higher (poor shielding by d and f electrons). So "gradually decreases from Si to Pb" is false.
Statement II: C is a non-metal, Si and Ge are metalloids, Sn and Pb are metals. True.
Answer: (D). | ||||
| 67 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements On heating to 513 K: $\mathrm{2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2}$.
Mn goes from +7 to +6 (manganate) and +4 ($\mathrm{MnO_2}$), while oxide oxygen is oxidised to $\mathrm{O_2}$.
Answer: (B). | ||||
| 68 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements In neutral or faintly alkaline solution permanganate is reduced to $\mathrm{MnO_2}$ and oxidises iodide to iodate:
$\mathrm{2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-}$.
(In acid solution iodide goes only to $\mathrm{I_2}$.)
Answer: (A). | ||||
| 69 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $\mathrm{Mn_2(CO)_{10}}$: two $\mathrm{Mn(CO)_5}$ units joined by a Mn–Mn bond only; all CO terminal.
$\mathrm{Co_2(CO)_8}$ (solid form): Co–Co bond with two CO groups bridging the two Co atoms and six terminal CO.
$\mathrm{Os_3(CO)_{12}}$ and $\mathrm{Ru_3(CO)_{12}}$: triangles of metal atoms with metal–metal bonds and only terminal CO (unlike $\mathrm{Fe_3(CO)_{12}}$, which has bridges).
Answer: (B). | ||||
| 70 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | Matching listEasy | |
A easy matching list question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds A. Ziegler(–Natta) catalyst: $\mathrm{TiCl_4}$ with triethylaluminium — titanium (IV).
B. Blood pigment, haemoglobin: iron (III).
C. Wilkinson's catalyst, $\mathrm{[RhCl(PPh_3)_3]}$: rhodium (I).
D. Vitamin $\mathrm{B_{12}}$ (cyanocobalamin): cobalt (II).
A→IV, B→III, C→I, D→II.
Answer: (A). | ||||
| 71 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques When both N and S are present, sodium fusion gives sodium thiocyanate: $\mathrm{Na + C + N + S \rightarrow NaSCN}$.
In the test for nitrogen (ferrous sulphate, then acid), the iron(III) formed gives the blood-red complex $\mathrm{[Fe(SCN)]^{2+}}$ instead of Prussian blue: $\mathrm{Fe^{3+} + SCN^- \rightarrow [Fe(SCN)]^{2+}}$.
N alone gives Prussian blue; S alone gives no colour here.
Answer: (C) N and S. | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Delocalisation through conjugation of a $\pi$ bond with a lone pair (or with another $\pi$ bond) on the adjacent atom is the resonance (mesomeric) effect, e.g. in aniline or vinyl chloride.
Electromeric effect is a temporary shift of $\pi$ electrons caused by an attacking reagent; hyperconjugation involves C–H $\sigma$ bonds; the inductive effect is polarisation through $\sigma$ bonds.
Answer: (B). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesElectronic effects | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on electronic effects, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The real molecule (the resonance hybrid) is more stable than any single canonical structure. The energy by which it lies below the most stable (lowest-energy) contributing structure is the resonance energy (resonance stabilisation energy).
Answer: (A). | ||||
| 74 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons With $\mathrm{Cl_2}$ in light ($h\nu$) the reaction is free-radical substitution at the allylic position (the allylic radical is resonance-stabilised), not addition: the C=C stays and Cl replaces allylic H. Product A keeps the ring double bond with Cl on the allylic carbon(s) — of the structures offered, the 3,6-dichlorocyclohexene.
In $\mathrm{CCl_4}$ in the dark, $\mathrm{Cl_2}$ adds across the double bond (via a chloronium ion, anti addition): product B is 1,2-dichlorocyclohexane.
Only option (A) has A as the allylic-substitution product and B as the addition product.
Answer: (A). | ||||
| 75 | Haloalkanes and HaloarenesNature of the C–X bond | Class 12: Haloalkanes and Haloarenes | Assertion and reasonEasy | |
A easy assertion and reason question on nature of the c–x bond, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes A: in phenol the C–O bond has partial double-bond character (the O lone pair is in resonance with the ring) and the carbon is $sp^2$, so the OH cannot be replaced by halogen the way it is in alcohols. Aryl halides are not made this way. A is true.
R: phenols do not react with halogen acids (HCl, HBr) to replace OH at all, let alone violently. R is false.
Answer: (C). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The starting compound is styrene (the ring is drawn as a hexagon with an inner circle, i.e. benzene) with a vinyl group.
(i) $\mathrm{H_2O/H^+}$: Markovnikov hydration through the benzylic carbocation gives 1-phenylethanol, $\mathrm{C_6H_5CH(OH)CH_3}$.
(ii) $\mathrm{CrO_3}$ oxidises the secondary alcohol to the ketone acetophenone, $\mathrm{C_6H_5COCH_3}$.
(iii) $\mathrm{H_2N{-}NH_2}$, KOH, heat: Wolff–Kishner reduction turns C=O into $\mathrm{CH_2}$: ethylbenzene, $\mathrm{C_6H_5CH_2CH_3}$.
The hydrazone (C) is only the intermediate of step (iii); with KOH and heat it loses $\mathrm{N_2}$.
Answer: (A). | ||||
| 77 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Two reactions occur with excess concentrated HBr on heating.
Ether cleavage: in an alkyl aryl ether, $\mathrm{Br^-}$ attacks the ethyl carbon (the aryl C–O bond has partial double-bond character and is not broken), giving the phenol and $\mathrm{CH_3CH_2Br}$. The phenolic OH is not replaced by Br.
Addition to the vinyl group: HBr adds by Markovnikov's rule through the stable benzylic carbocation, giving $\mathrm{-CH(Br)CH_3}$ (no peroxide is present, so no anti-Markovnikov addition).
Product P: 3-(1-bromoethyl)phenol.
Answer: (C). | ||||
| 78 | BiomoleculesProteins | Class 12: Biomolecules | Single correctEasy | |
A easy single correct question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Proteins are polymers of $\alpha$-amino acids, $\mathrm{H_2N{-}CH(R){-}COOH}$, in which the amino group is on the carbon next to the carboxyl group. Hydrolysis of the peptide bonds gives back these $\alpha$-amino acids.
Answer: (A). | ||||
| 79 | Principles of Qualitative AnalysisAnalysis of anions | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on analysis of anions, from Principles of Qualitative Analysis. NCERT: No matching NCERT chapter The yellow solution is sodium chromate; with acid and $\mathrm{H_2O_2}$ it gives blue $\mathrm{CrO_5}$.
$\mathrm{CrO_5}$ has the butterfly structure: one Cr=O oxygen and two peroxo ($\mathrm{O_2^{2-}}$) groups.
Oxidation state: $x + (-2) + 2(-2) = 0 \Rightarrow x = +6$. (Treating all five O as $-2$ would give an impossible +10.)
Answer: (B). | ||||
| 80 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{NH_2}$ is a strongly activating ortho/para director: $\mathrm{Br^+}$ attacks only ortho and para, giving 2,4,6-tribromoaniline.
(A) attack at the ortho carbon, the charge delocalised onto nitrogen (iminium form) — involved.
(B) attack at the para carbon, charge on nitrogen — involved.
(D) attack at the ortho carbon (C2), charge on C3, which is ortho to the attacked carbon — a valid contributor of ortho attack — involved.
(C) the $\mathrm{sp^3}$ carbon carrying H and Br is C3, meta to $\mathrm{NH_2}$: this arenium ion comes from meta attack, which does not occur (its charge can never reach the nitrogen).
Answer: (C). | ||||
| 81 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 L of solution: mass $= 1000\times 1.25 = 1250$ g, of which $\mathrm{H_2SO_4}$ is $0.314\times 1250 = 392.5$ g.
Moles $= \frac{392.5}{98} = 4.005$.
Molarity $\approx 4.0$ M, nearest integer 4.
(Formula: $M = \frac{10\times\%\times d}{M_w} = \frac{10\times 31.4\times 1.25}{98} = 4.005$.)
Answer: 4. | ||||
| 82 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalMedium | |
A medium numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Molecular orbital configurations (bond order, unpaired electrons):
$\mathrm{H_2}$ (2 e): $\sigma1s^2$ — BO 1, diamagnetic.
$\mathrm{He_2^+}$ (3 e): BO 0.5, paramagnetic.
$\mathrm{O_2^+}$ (15 e): BO 2.5, paramagnetic.
$\mathrm{N_2^{2-}}$ (16 e, like $\mathrm{O_2}$): BO 2, paramagnetic.
$\mathrm{O_2^{2-}}$ (18 e): BO 1, diamagnetic.
$\mathrm{F_2}$ (18 e): BO 1, diamagnetic.
$\mathrm{Ne_2^+}$ (19 e): BO 0.5, paramagnetic.
$\mathrm{B_2}$ (10 e): $\ldots\pi2p_x^1\pi2p_y^1$ — BO 1, two unpaired electrons, paramagnetic.
Only $\mathrm{B_2}$ is both paramagnetic and of bond order one.
Answer: 1. | ||||
| 83 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Lone pairs on the central atom:
$\mathrm{O_3}$: central O has 1 lone pair (bent). Yes.
$\mathrm{H_2O}$: 2. No.
$\mathrm{SF_4}$: 1 (see-saw). Yes.
$\mathrm{ClF_3}$: 2 (T-shaped). No.
$\mathrm{NH_3}$: 1 (pyramidal). Yes.
$\mathrm{BrF_5}$: 1 (square pyramidal). Yes.
$\mathrm{XeF_4}$: 2 (square planar). No.
Count: 4.
Answer: 4. | ||||
| 84 | SolutionsColligative properties | Class 12: Solutions | NumericalEasy | |
A easy numerical question on colligative properties, from Solutions. NCERT: Class 12: Solutions $\pi = CRT$ with the same concentration, so $\pi \propto T$.
$\pi_2 = 7\times 10^5\times\frac{283}{273} = 7.2564\times 10^5$ Pa $= 72.56\times 10^4$ N m$^{-2}$.
Nearest integer: 73.
Answer: 73. | ||||
| 85 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium $K_p = K_c(RT)^{\Delta n_g}$ with $\Delta n_g = 2 - 1 = 1$.
$K_c = \frac{K_p}{RT} = \frac{0.492}{0.082\times 300} = \frac{0.492}{24.6} = 0.02 = 2\times 10^{-2}$ mol L$^{-1}$.
Answer: 2. | ||||
| 86 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | NumericalEasy | |
A easy numerical question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Charge $Q = It = 0.015\times 15\times 60 = 13.5$ C.
$\mathrm{Zn^{2+} + 2e^- \rightarrow Zn}$: 2 F deposit 65.4 g.
Mass $= \frac{13.5\times 65.4}{2\times 96500} = \frac{882.9}{193000} = 4.575\times 10^{-3}$ g $= 45.75\times 10^{-4}$ g.
Nearest integer: 46.
Answer: 46. | ||||
| 87 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | NumericalMedium | |
A medium numerical question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Write each constant in Arrhenius form, $K_i = A_ie^{-E_{a_i}/RT}$:
$K = \frac{A_1A_2}{A_3}\,e^{-(E_{a_1} + E_{a_2} - E_{a_3})/RT}$.
So the overall activation energy is $E_a = E_{a_1} + E_{a_2} - E_{a_3} = 40 + 50 - 60 = 30$ kJ/mol.
Answer: 30. | ||||
| 88 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalEasy | |
A easy numerical question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Furan: O in the ring. Thiophene: S in the ring. Pyridine and pyrrole: N. Tyrosine: $\mathrm{HO{-}C_6H_4{-}CH_2CH(NH_2)COOH}$, no S.
Cysteine: $\mathrm{HS{-}CH_2{-}CH(NH_2){-}COOH}$ contains sulphur (a heteroatom in the organic sense: any atom other than C and H).
Compounds containing S: thiophene and cysteine — 2.
Answer: 2. | ||||
| 89 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalEasy | |
A easy numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The alkene drawn is trans-but-2-ene, $\mathrm{CH_3CH{=}CHCH_3}$.
Reductive ozonolysis ($\mathrm{O_3}$, then $\mathrm{Zn/H_2O}$) cleaves the C=C and puts a C=O on each carbon: $\mathrm{CH_3CH{=}CHCH_3 \rightarrow 2\,CH_3CHO}$.
Both halves are the same, so P is acetaldehyde (ethanal), $\mathrm{CH_3CHO}$, with 1 oxygen atom per molecule.
Answer: 1. | ||||
| 90 | Aldehydes, Ketones and Carboxylic AcidsTests to distinguish aldehydes and ketones | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalEasy | |
A easy numerical question on tests to distinguish aldehydes and ketones, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Fehling's solution (Cu(II) tartrate complex) is reduced to red $\mathrm{Cu_2O}$ by aliphatic aldehydes, but not by aromatic aldehydes or by ketones.
Positive: acetaldehyde, methanal, cyclohexanecarbaldehyde (the CHO is on an $sp^3$ carbon, so it behaves as an aliphatic aldehyde).
Negative: benzaldehyde and 4-nitrobenzaldehyde (aromatic aldehydes), acetone and acetophenone (ketones).
Count: 3.
Answer: 3. | ||||
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